5
$\begingroup$

Compact surfaces are classified up to homeomorphism by their fundamental group (defined up to isomorphism). However, 2 compact Riemann surfaces may well be homeomorphic but not conformally equivalent. Still, they are related to their fundamental groups by the fact that the universal cover map is holomorphic : so for say a compact hyperbolic Riemann surface $M$, it can be described as the quotient of the upper half plane under the action of a Fuchsian group which is isomorphic to the fundamental group. That Fuchsian group is not unique though.

My question is this : what is the notion of equivalence needed for two Fuchsian groups to generate the same (hyperbolic) Riemann surface ? It needs a priori to be stronger than just isomorphism.

$\endgroup$
1
$\begingroup$

Unless I'm overlooking something:

The groups must be conjugate in $\operatorname{Aut}(\mathbb{H})$.

Say we have two hyperbolic surfaces $S,\, T$ with their universal covers $\pi_S,\, \pi_T$ and a biholomorhism $\varphi \colon S \to T$. Choosing a point $s \in S$ and $\sigma \in \mathbb{H}$ above $s$, and a point $\tau \in \mathbb{H}$ above $t = \varphi(s)$, there is a unique lift of $\varphi$ to $\Phi \colon \mathbb{H} \to \mathbb{H}$ with $\Phi(\sigma) = \tau$ such that the diagram

$$\begin{matrix} \hphantom{XY}\mathbb{H} & \overset{\Phi}{\longrightarrow} & \mathbb{H}\hphantom{XY}\\ \pi_S\downarrow & & \downarrow\pi_T\\ \hphantom{XY}S & \underset{\varphi}{\longrightarrow} & T\hphantom{XY} \end{matrix}$$

is commutative. Now let $\delta$ a deck transformation of $\pi_S$. Then $\pi_T \circ \Phi \circ \delta = \varphi \circ \pi_s \circ \delta = \varphi \circ \pi_S$, hence $\Phi \circ \delta$ is the unique lift of $\varphi$ that maps $\sigma$ to $\Phi(\delta(\sigma))$. Let $\hat{\delta}$ the deck transformation of $\pi_T$ that maps $\tau$ to $\Phi(\delta(\sigma))$. Then we see that $\Phi \circ \delta = \hat{\delta} \circ \Phi$, or $\hat{\delta} = \Phi \circ \delta \circ \Phi^{-1}$. That is, conjugation with $\Phi$ gives an isomorphism $\operatorname{Deck}(\pi_S) \to \operatorname{Deck}(\pi_T)$.

Conversely, if we have $\operatorname{Deck}(\pi_T) = \Phi\operatorname{Deck}(\pi_S) \Phi^{-1}$ for a $\Phi \in \operatorname{Aut}(\mathbb{H})$, we can push down $\Phi$ to a biholomorphism $\varphi$ between $S$ and $T$.

$\endgroup$
  • $\begingroup$ thanks ! I guess what kind of confused me was that two isomorphic groups need not be conjugated in $\mathrm{Aut}(\mathbb{H})$... It's clearer for me now ! $\endgroup$ – Khal665 Jul 12 '13 at 22:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.